Đáp án:

\(\begin{array}{l}
7,\\
\left[ \begin{array}{l}
x = 1\\
x =  - 4\\
x =  - 1\\
x =  - 2
\end{array} \right.\\
8,\\
\left[ \begin{array}{l}
x = 1\\
x =  - 3\\
x =  - 1
\end{array} \right.\\
9,\\
\left[ \begin{array}{l}
x = 1\\
x =  - 2
\end{array} \right.\\
10,\\
\left[ \begin{array}{l}
x = 1\\
x =  - 9
\end{array} \right.\\
11,\\
\left[ \begin{array}{l}
x = 1\\
x =  - 11
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
7,\\
\left( {{x^2} + 3x + 1} \right)\left( {{x^2} + 3x - 3} \right) = 5\\
 \Leftrightarrow \left[ {\left( {{x^2} + 3x - 1} \right) + 2} \right].\left[ {\left( {{x^2} + 3x - 1} \right) - 2} \right] = 5\\
 \Leftrightarrow {\left( {{x^2} + 3x - 1} \right)^2} - {2^2} = 5\\
 \Leftrightarrow {\left( {{x^2} + 3x - 1} \right)^2} - 4 - 5 = 0\\
 \Leftrightarrow {\left( {{x^2} + 3x - 1} \right)^2} - 9 = 0\\
 \Leftrightarrow {\left( {{x^2} + 3x - 1} \right)^2} - {3^2} = 0\\
 \Leftrightarrow \left[ {\left( {{x^2} + 3x - 1} \right) - 3} \right].\left[ {\left( {{x^2} + 3x - 1} \right) + 3} \right] = 0\\
 \Leftrightarrow \left( {{x^2} + 3x - 4} \right).\left( {{x^2} + 3x + 2} \right) = 0\\
 \Leftrightarrow \left[ {\left( {{x^2} - x} \right) + \left( {4x - 4} \right)} \right].\left[ {\left( {{x^2} + x} \right) + \left( {2x + 2} \right)} \right] = 0\\
 \Leftrightarrow \left[ {x.\left( {x - 1} \right) + 4.\left( {x - 1} \right)} \right].\left[ {x\left( {x + 1} \right) + 2.\left( {x + 1} \right)} \right] = 0\\
 \Leftrightarrow \left( {x - 1} \right)\left( {x + 4} \right).\left( {x + 1} \right)\left( {x + 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
x + 4 = 0\\
x + 1 = 0\\
x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x =  - 4\\
x =  - 1\\
x =  - 2
\end{array} \right.\\
8,\\
{\left( {{x^2} + 2x} \right)^2} - 2{x^2} - 4x = 3\\
 \Leftrightarrow {\left( {{x^2} + 2x} \right)^2} - \left( {2{x^2} + 4x} \right) = 3\\
 \Leftrightarrow {\left( {{x^2} + 2x} \right)^2} - 2.\left( {{x^2} + 2x} \right) - 3 = 0\\
 \Leftrightarrow {\left( {{x^2} + 2x} \right)^2} - 3\left( {{x^2} + 2x} \right) + \left( {{x^2} + 2x} \right) - 3 = 0\\
 \Leftrightarrow \left[ {{{\left( {{x^2} + 2x} \right)}^2} - 3.\left( {{x^2} + 2x} \right)} \right] + \left[ {\left( {{x^2} + 2x} \right) - 3} \right] = 0\\
 \Leftrightarrow \left( {{x^2} + 2x} \right).\left[ {\left( {{x^2} + 2x} \right) - 3} \right] + \left[ {\left( {{x^2} + 2x} \right) - 3} \right] = 0\\
 \Leftrightarrow \left( {{x^2} + 2x} \right).\left( {{x^2} + 2x - 3} \right) + \left( {{x^2} + 2x - 3} \right) = 0\\
 \Leftrightarrow \left( {{x^2} + 2x - 3} \right).\left[ {\left( {{x^2} + 2x} \right) + 1} \right] = 0\\
 \Leftrightarrow \left( {{x^2} + 2x - 3} \right).\left( {{x^2} + 2x + 1} \right) = 0\\
 \Leftrightarrow \left[ {\left( {{x^2} - x} \right) + \left( {3x - 3} \right)} \right].\left( {{x^2} + 2.x.1 + {1^2}} \right) = 0\\
 \Leftrightarrow \left[ {x.\left( {x - 1} \right) + 3.\left( {x - 1} \right)} \right].{\left( {x + 1} \right)^2} = 0\\
 \Leftrightarrow \left( {x - 1} \right)\left( {x + 3} \right).{\left( {x + 1} \right)^2} = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
x + 3 = 0\\
x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x =  - 3\\
x =  - 1
\end{array} \right.\\
9,\\
{\left( {{x^2} + x} \right)^2} - \left( {{x^2} + x} \right) - 2 = 0\\
 \Leftrightarrow {\left( {{x^2} + x} \right)^2} - 2\left( {{x^2} + x} \right) + \left( {{x^2} + x} \right) - 2 = 0\\
 \Leftrightarrow \left[ {{{\left( {{x^2} + x} \right)}^2} - 2.\left( {{x^2} + x} \right)} \right] + \left[ {\left( {{x^2} + x} \right) - 2} \right] = 0\\
 \Leftrightarrow \left( {{x^2} + x} \right).\left[ {\left( {{x^2} + x} \right) - 2} \right] + \left[ {\left( {{x^2} + x} \right) - 2} \right] = 0\\
 \Leftrightarrow \left( {{x^2} + x} \right).\left( {{x^2} + x - 2} \right) + \left( {{x^2} + x - 2} \right) = 0\\
 \Leftrightarrow \left( {{x^2} + x - 2} \right).\left[ {\left( {{x^2} + x} \right) + 1} \right] = 0\\
 \Leftrightarrow \left( {{x^2} + x - 2} \right)\left( {{x^2} + x + 1} \right) = 0\\
 \Leftrightarrow \left[ {\left( {{x^2} - x} \right) + \left( {2x - 2} \right)} \right].\left[ {\left( {{x^2} + x + \dfrac{1}{4}} \right) + \dfrac{3}{4}} \right] = 0\\
 \Leftrightarrow \left[ {x.\left( {x - 1} \right) + 2.\left( {x - 1} \right)} \right].\left[ {\left( {{x^2} + 2.x.\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right) + \dfrac{3}{4}} \right] = 0\\
 \Leftrightarrow \left( {x - 1} \right)\left( {x + 2} \right).\left[ {{{\left( {x + \dfrac{1}{2}} \right)}^2} + \dfrac{3}{4}} \right] = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
x + 2 = 0\\
{\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x =  - 2\\
{\left( {x + \dfrac{1}{2}} \right)^2} =  - \dfrac{3}{4}\,\,\,\left( L \right)
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x =  - 2
\end{array} \right.\\
10,\\
\left( {x + 1} \right)\left( {x + 3} \right)\left( {x + 5} \right)\left( {x + 7} \right) = 384\\
 \Leftrightarrow \left[ {\left( {x + 1} \right)\left( {x + 7} \right)} \right].\left[ {\left( {x + 3} \right)\left( {x + 5} \right)} \right] = 384\\
 \Leftrightarrow \left( {{x^2} + 7x + x + 7} \right)\left( {{x^2} + 5x + 3x + 15} \right) = 384\\
 \Leftrightarrow \left( {{x^2} + 8x + 7} \right)\left( {{x^2} + 8x + 15} \right) = 384\\
 \Leftrightarrow \left[ {\left( {{x^2} + 8x + 11} \right) - 4} \right].\left[ {\left( {{x^2} + 8x + 11} \right) + 4} \right] - 384 = 0\\
 \Leftrightarrow {\left( {{x^2} + 8x + 11} \right)^2} - {4^2} - 384 = 0\\
 \Leftrightarrow {\left( {{x^2} + 8x + 11} \right)^2} - 16 - 384 = 0\\
 \Leftrightarrow {\left( {{x^2} + 8x + 11} \right)^2} - 400 = 0\\
 \Leftrightarrow {\left( {{x^2} + 8x + 11} \right)^2} - {20^2} = 0\\
 \Leftrightarrow \left[ {\left( {{x^2} + 8x + 11} \right) - 20} \right].\left[ {\left( {{x^2} + 8x + 11} \right) + 20} \right] = 0\\
 \Leftrightarrow \left( {{x^2} + 8x - 9} \right).\left( {{x^2} + 8x + 31} \right) = 0\\
 \Leftrightarrow \left[ {\left( {{x^2} - x} \right) + \left( {9x - 9} \right)} \right].\left[ {\left( {{x^2} + 8x + 16} \right) + 15} \right] = 0\\
 \Leftrightarrow \left[ {x.\left( {x - 1} \right) + 9.\left( {x - 1} \right)} \right].\left[ {\left( {{x^2} + 2.x.4 + {4^2}} \right) + 15} \right] = 0\\
 \Leftrightarrow \left( {x - 1} \right)\left( {x + 9} \right).\left[ {{{\left( {x + 4} \right)}^2} + 15} \right] = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
x + 9 = 0\\
{\left( {x + 4} \right)^2} + 15 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x =  - 9\\
{\left( {x + 4} \right)^2} =  - 15\,\,\,\left( L \right)
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x =  - 9
\end{array} \right.\\
11,\\
x\left( {x + 4} \right)\left( {x + 6} \right)\left( {x + 10} \right) - 385 = 0\\
 \Leftrightarrow \left[ {x.\left( {x + 10} \right)} \right].\left[ {\left( {x + 4} \right)\left( {x + 6} \right)} \right] - 385 = 0\\
 \Leftrightarrow \left( {{x^2} + 10x} \right).\left( {{x^2} + 6x + 4x + 24} \right) - 385 = 0\\
 \Leftrightarrow \left( {{x^2} + 10x} \right)\left( {{x^2} + 10x + 24} \right) - 385 = 0\\
 \Leftrightarrow \left[ {\left( {{x^2} + 10x + 12} \right) - 12} \right].\left[ {\left( {{x^2} + 10x + 12} \right) + 12} \right] - 385 = 0\\
 \Leftrightarrow {\left( {{x^2} + 10x + 12} \right)^2} - {12^2} - 385 = 0\\
 \Leftrightarrow {\left( {{x^2} + 10x + 12} \right)^2} - 144 - 385 = 0\\
 \Leftrightarrow {\left( {{x^2} + 10x + 12} \right)^2} - 529 = 0\\
 \Leftrightarrow {\left( {{x^2} + 10x + 12} \right)^2} - {23^2} = 0\\
 \Leftrightarrow \left[ {\left( {{x^2} + 10x + 12} \right) - 23} \right].\left[ {\left( {{x^2} + 10x + 12} \right) + 23} \right] = 0\\
 \Leftrightarrow \left( {{x^2} + 10x - 11} \right).\left( {{x^2} + 10x + 35} \right) = 0\\
 \Leftrightarrow \left[ {\left( {{x^2} - x} \right) + \left( {11x - 11} \right)} \right].\left[ {\left( {{x^2} + 10x + 25} \right) + 10} \right] = 0\\
 \Leftrightarrow \left[ {x.\left( {x - 1} \right) + 11.\left( {x - 1} \right)} \right].\left[ {\left( {{x^2} + 2.x.5 + {5^2}} \right) + 10} \right] = 0\\
 \Leftrightarrow \left( {x - 1} \right)\left( {x + 11} \right).\left[ {{{\left( {x + 5} \right)}^2} + 10} \right] = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
x + 11 = 0\\
{\left( {x + 5} \right)^2} + 10 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x =  - 11\\
{\left( {x + 5} \right)^2} =  - 10\,\,\,\,\left( L \right)
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x =  - 11
\end{array} \right.
\end{array}\)